Supremum of the function $f(x,y,z,u)=\frac{x(1-x)y(1-y)z(1-u)}{1-(1-xy)z}$

Question

Consider a function $$f(x,y,z,u)=\frac{x(1-x)y(1-y)z(1-u)}{1-(1-xy)z} $$ where $x,y,z,u\in (0,1)$

I need the supremum of $f(x,y,z,u)$ for $x,y,z,u\in (0,1)$

We have $1-(1-xy)z=1-z+xyz$

Now since $x,y,z\in(0,1)$, so we have $1-z>0$ and by AM-GM inequality $$1-(1-xy)z=1-z+xyz\geq 2\sqrt{(1-z)xyz}$$ So we have $$f(x,y,z,u)\leq \frac{1}{2} \frac{\sqrt{x}(1-x)\sqrt{y}(1-y)z(1-u)}{\sqrt{z(1-z)}} $$ Now if $G(x)=\sqrt{x}(1-x)$ and we put $t=\sqrt{x}$ then the supremum of $G(x)=t(1-t^2)$ occurs when $1-3t^3=0$ so at $t=\frac{1}{\sqrt{3}}$ so we have $$f(x,y,z,u)<\frac{1}{2}(\frac{\frac{1}{\sqrt{3}}(1-\frac{1}{3})\frac{1}{\sqrt{3}}(1-\frac{1}{3})z(1-u)}{\sqrt{z(1-z)}}) $$ So $$f(x,y,z,u)<\frac{2}{27}\frac{z(1-u)}{\sqrt{z(1-z)}}$$ Please solve this question.

Answer 1

Writing the function as $$ f(x,y,z,u) = \frac{xyz}{1-z+xyz}(1-x)(1-y)(1-u), $$ we have a product of four terms all strictly less than $1$. Therefore $f(x,y,z,u) < 1$ for all $x,y,z,u\in(0,1)$.

To show that $1$ is the supremum, we need to show we can get arbitrarily close to it. The product form above suggests making $x, y, u$ close to $0$ and $z$ close to $1$. So we consider $$ f(x,y,1-w,u) = \frac{(1-x)(1-y)(1-u)}{1 + \frac{w}{xy(1-w)}} > 1 - \left[x+y+u+\frac{w}{xy(1-w)}\right] $$ The term in brackets can be made arbitrarily small through suitable choice of $x,y,u,w\in (0,1)$, so $1$ is indeed the supremum.

Answer 2

Here

$$ f(x,y,z,u)=\frac{x(1-x)y(1-y)z(1-u)}{1-(1-xy)z} = g(x,y,z)(1-u) $$

now as $g(x,y,z)\ge 0$ we have that $u=0$. Considering now $\epsilon> 0$, $0\lt z+\epsilon\lt 1$ we have

$$ g(x,y,z+\epsilon) =\frac{x(1-x)y(1-y)z}{1-(1-xy)(z+\epsilon)}+\frac{x(1-x)y(1-y)\epsilon}{1-(1-xy)(z+\epsilon)}\gt \frac{x(1-x)y(1-y)z}{1-(1-xy)z} $$

because

$$ \frac{x(1-x)y(1-y)z}{1-(1-xy)(z+\epsilon)}\gt \frac{x(1-x)y(1-y)z}{1-(1-xy)z} $$

so the supremum is attained at

$$ \sup f(x,y,z,u) = \lim_{\cases{x(z)\to 0\\ y(z)\to 0\\ z\to 1}}f(x,y,z,0)=1 $$

NOTE

To determine a possible path $x(z),y(z)$ we follow with the problem

$$ \max_{s}u(s,z) = \frac{s^2(1-s)^2z}{1-(1-s^2)z},\ \ \ (\text{here}\ \ s=x=y) $$

now from $u'(s,z) = 0$ and solving a cubic we got a real root as

$$ s(z) = \frac{\sqrt[3]{2} \left(\sqrt{3} \sqrt{(1-z)^2 z^3 (32-5 z)}+9 (1-z) z^2\right)^{2/3}-4 \sqrt[3]{3} (1-z) z}{6^{2/3} z \sqrt[3]{\sqrt{3} \sqrt{(1-z)^2 z^3 (32-5 z)}+9 (1-z) z^2}} $$

Follows a plot for $s(z), \ 0 < z < 1$

and also for $u(s(z),z), \ \ 0 < z < 1$

Answer 3

Due to symmetry we can take $x=y$ (Still trying to find a better justification) and as eyeballfrog commented $u\rightarrow 0$ gives supremum. Hence, we consider the function $f(x,z)=\frac{x^2(1-x)^2z}{1-(1-x^2)z}$ on $D=(0,1)×(0,1)$. Assume that $f(x,z)>1$. Then,$z>\frac{1}{1-x^2(1-(1-x)^2)}>1$ when $0<x<1$. Therefore, $\sup_D f(x,z)=1$, achived by $x\rightarrow 0$, $z\rightarrow 1$.

Answer 4

Previous version of this question was deleted and only today I found this one. May be this is not complete answer, but let me bring here some calculation in hope, they will be useful: First derivatives are: $$\begin{cases}\dfrac{\partial \:}{\partial \:x}\left(\dfrac{x\left(1-x\right)y\left(1-y\right)z\left(1-u\right)}{1-\left(1-xy\right)z}\right)=\dfrac{yz\left(-x^2yz+2xz-2x-z+1\right)\left(1-y\right)\left(1-u\right)}{\left(1-z\left(1-xy\right)\right)^2}\\ \dfrac{\partial \:}{\partial \:y}\left(\dfrac{x\left(1-x\right)y\left(1-y\right)z\left(1-u\right)}{1-\left(1-xy\right)z}\right)=\dfrac{xz\left(-xy^2z+2yz-2y-z+1\right)\left(1-x\right)\left(1-u\right)}{\left(1-z\left(1-xy\right)\right)^2}\\ \dfrac{\partial \:}{\partial \:z}\left(\dfrac{x\left(1-x\right)y\left(1-y\right)z\left(1-u\right)}{1-\left(1-xy\right)z}\right)=\dfrac{xy\left(1-x\right)\left(1-y\right)\left(1-u\right)}{\left(1-z\left(1-xy\right)\right)^2}\\ \dfrac{\partial \:}{\partial \:u}\left(\dfrac{x\left(1-x\right)y\left(1-y\right)z\left(1-u\right)}{1-\left(1-xy\right)z}\right)=-\dfrac{xyz\left(1-x\right)\left(1-y\right)}{1-z\left(1-xy\right)}\\ \end{cases}$$ From last two lines is obvious, that extremum cannot be inside $(0,1)^4$.

Calculating Hessian matrix (for saving space I am not including explicit formulas for second derivatives) $$\begin{vmatrix} \dfrac{\partial^2 }{\partial x^2}f & \dfrac{\partial^2 }{\partial y \partial x}f & \dfrac{\partial^2 }{\partial z \partial x}f & \dfrac{\partial^2 }{\partial u \partial x}f\\ \dfrac{\partial^2 }{\partial x \partial y}f & \dfrac{\partial^2 }{ \partial y^2}f & \dfrac{\partial^2 }{ \partial z \partial y}f & \dfrac{\partial^2 }{ \partial u \partial y}f\\ \dfrac{\partial^2 }{\partial x \partial z}f & \dfrac{\partial^2 }{ \partial y \partial z}f & \dfrac{\partial^2 }{ \partial^2 z }f & \dfrac{\partial^2 }{ \partial u \partial z }f\\ \dfrac{\partial^2 }{\partial x \partial u}f & \dfrac{\partial^2 }{ \partial y \partial u}f & \dfrac{\partial^2 }{ \partial z \partial u}f & \dfrac{\partial^2 }{ \partial^2 u }f \end{vmatrix}_{\Big|_{(1,1,z,u)}}=\begin{vmatrix} 0 & z(1-u) & 0 & 0\\ z(1-u) & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{vmatrix}=0$$

shows, that Second partial derivative test is not applicable in points $(1,1,z,u)$. I'll try more deep methods and bring them in nearest future.